Solution -「CF 757F」Team Rocket Rises Again

$\mathcal{Description}$

  link.
  给定 $n$ 个点 $m$ 条边的无向图和一个源点 $s$。要求删除一个不同与 $s$ 的结点 $u$，使得有最多的点到 $s$ 的最短距离改变。求出此时最短距离改变的结点的数量。
  $n\le2\times10^5,m\le3\times10^5$。

$\mathcal{Solution}$

  首先，以 $s$ 为源点跑一个单源最短路。设 $s$ 到 $u$ 的距离为 $dist_u$。
  接着枚举所有点 $u$ 与其一条边 $(u,v)$。若满足 $dist_u+\operatorname{cost}(u,v)=dist_v$，则表示该边是 $v$ 最短路径的一条转移边，将其加入新图 $G$ 中。
  显然 $G$ 是有向无环图，所以直接建立支配树，求出子树大小最大的结点即可。

$\mathcal{Code}$

#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>

#define cost first
#define node second
#define adj( g, u, v ) \
	for ( int _eid = g.head[u], v; v = g.to[_eid], _eid; _eid = g.nxt[_eid] )

typedef long long LL;
typedef std::pair<LL, int> pli;

const int MAXN = 2e5, MAXM = 6e5, MAXLG = 17;
int n, m, s, dep[MAXN + 5], siz[MAXN + 5], rnk[MAXN + 5], fa[MAXN + 5][MAXLG + 5];
LL dist[MAXN + 5];
std::queue<int> que;
std::vector<int> pre[MAXN + 5];
std::vector<pli> sour[MAXN + 5];

inline int rint () {
	int x = 0; char s = getchar ();
	for ( ; s < '0' || '9' < s; s = getchar () );
	for ( ; '0' <= s && s <= '9'; s = getchar () ) x = x * 10 + ( s ^ '0' );
	return x;
}

struct Graph {
	int ecnt, head[MAXN + 5], to[MAXM + 5], nxt[MAXM + 5], ind[MAXN + 5];
	inline void link ( const int s, const int t ) {
		++ ind[to[++ ecnt] = t], nxt[ecnt] = head[s], head[s] = ecnt;
		pre[t].push_back ( s );
	}
} dag, domt;

inline int LCA ( int u, int v ) {
	if ( dep[u] < dep[v] ) u ^= v ^= u ^= v;
	for ( int i = 17; ~ i; -- i ) if ( dep[fa[u][i]] >= dep[v] ) u = fa[u][i];
	if ( u == v ) return u;
	for ( int i = 17; ~ i; -- i ) if ( fa[u][i] ^ fa[v][i] ) u = fa[u][i], v = fa[v][i];
	return fa[u][0];
}

inline void calc ( const int u ) {
	siz[u] = 1;
	adj ( domt, u, v ) calc ( v ), siz[u] += siz[v];
}

inline void Dijkstra ( const int s ) {
	static bool vis[MAXN + 5] {};
	static std::priority_queue<pli, std::vector<pli>, std::greater<pli> > pque;
	memset ( dist, 0x3f, sizeof dist ), pque.push ( { dist[s] = 0, s } );
	while ( ! pque.empty () ) {
		pli p = pque.top (); pque.pop ();
		if ( vis[p.node] ) continue;
		vis[p.node] = true;
		for ( pli e: sour[p.node] ) {
			if ( ! vis[e.node] && p.cost + e.cost < dist[e.node] ) {
				pque.push ( { dist[e.node] = p.cost + e.cost, e.node } );
			}
		}
	}
}

int main () {
	n = rint (), m = rint (), s = rint ();
	for ( int i = 1, u, v, w; i <= m; ++ i ) {
		u = rint (), v = rint (), w = rint ();
		sour[u].push_back ( { LL ( w ), v } );
		sour[v].push_back ( { LL ( w ), u } );
	}
	Dijkstra ( s );
	for ( int i = 1; i <= n; ++ i ) {
		for ( pli e: sour[i] ) {
			if ( dist[e.node] == dist[i] + e.cost ) {
				dag.link ( i, e.node );
			}
		}
	}
	que.push ( s );
	int cnt = 0;
	for ( int u; ! que.empty (); que.pop () ) {
		rnk[++ cnt] = u = que.front ();
		adj ( dag, u, v ) if ( ! -- dag.ind[v] ) que.push ( v );
	}
	for ( int i = 1; i <= cnt; ++ i ) {
		int u = rnk[i], f = 0;
		if ( ! pre[u].empty () ) f = pre[u][0];
		for ( int j = 1; j < ( int ) pre[u].size (); ++ j ) f = LCA ( f, pre[u][j] );
		dep[u] = dep[fa[u][0] = f] + 1, domt.link ( f, u );
		for ( int j = 1; j <= 17; ++ j ) fa[u][j] = fa[fa[u][j - 1]][j - 1];
	}
	calc ( s );
	int ans = 0;
	for ( int i = 1; i <= n; ++ i ) if ( i ^ s ) ans = ans < siz[i] ? siz[i] : ans;
	printf ( "%d\n", ans );
	return 0;
}